将给定字符串的所有元音替换为单个元音的成本最小化

给定一个字符串S，任务是找到将给定字符串中的所有元音{a, e, i, o, u}转换为其中任何一个的最小成本。转换元音的成本由 ASCII 值的变化给出。
例子：

Input: S = “geeksforgeeks” 
Output: 10 
Explanation: 
Count of e’s = 4 
Count of o’s = 1 
Conversion from ‘o’ to ‘e’ costs abs(‘o’ – ‘e’) = 10. 
Hence, the output is 10.
Input: S = “coding” 
Output: 6 
Explanation: 
Conversion from ‘o’ to ‘i’ costs abs(‘o’ – ‘i’) = 6. 
Hence, the output is 10. 
 

方法：
这个想法是计算将字符串的每个元音转换为元音{a, e, i, o, u}之一的单独成本，并选择成本最低的元音。请按照以下步骤操作：

• 初始化 5 个变量cA、cE、cI、 cO和cU ，代价为0 ，分别存储元音{a, e, i, o, u}的代价。
• 遍历字符串，并为每个元音找到将其转换为所有其他元音的成本，并将其分别存储在变量cA、cE、cI、 cO和cU 中。
• 将所有元音转换为单个元音的最小成本由min(cA, cE, cI, cO, cU) 给出。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function that return true if the
// given character is a vowel
bool isVowel(char ch)
{
    if (ch == 'a' or ch == 'e'
        or ch == 'i' or ch == 'o'
        or ch == 'u')
        return true;
    else
        return false;
}
 
// Function to return the minimum
// cost to convert all the vowels
// of a string to a single one
int minCost(string S)
{
    // Stores count of
    // respective vowels
    int cA = 0;
    int cE = 0;
    int cI = 0;
    int cO = 0;
    int cU = 0;
 
    // Iterate through the string
    for (int i = 0; i < S.size(); i++) {
 
        // If a vowel is encountered
        if (isVowel(S[i])) {
 
            // Calculate the cost
            cA += abs(S[i] - 'a');
            cE += abs(S[i] - 'e');
            cI += abs(S[i] - 'i');
            cO += abs(S[i] - 'o');
            cU += abs(S[i] - 'u');
        }
    }
 
    // Return the minimum cost
    return min(min(min(min(cA, cE),
                    cI),
                cO),
            cU);
}
 
// Driver Code
int main()
{
    string S = "geeksforgeeks";
 
    cout << minCost(S) << endl;
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function that return true if the
// given character is a vowel
static boolean isVowel(char ch)
{
    if (ch == 'a' || ch == 'e' ||
        ch == 'i' || ch == 'o' ||
        ch == 'u')
        return true;
    else
        return false;
}
 
// Function to return the minimum
// cost to convert all the vowels
// of a string to a single one
static int minCost(String S)
{
     
    // Stores count of
    // respective vowels
    int cA = 0;
    int cE = 0;
    int cI = 0;
    int cO = 0;
    int cU = 0;
 
    // Iterate through the string
    for(int i = 0; i < S.length(); i++)
    {
         
        // If a vowel is encountered
        if (isVowel(S.charAt(i)))
        {
             
            // Calculate the cost
            cA += Math.abs(S.charAt(i) - 'a');
            cE += Math.abs(S.charAt(i) - 'e');
            cI += Math.abs(S.charAt(i) - 'i');
            cO += Math.abs(S.charAt(i) - 'o');
            cU += Math.abs(S.charAt(i) - 'u');
        }
    }
 
    // Return the minimum cost
    return Math.min(Math.min(Math.min(Math.min(cA, cE),
                                         cI), cO), cU);
}
 
// Driver code
public static void main(String[] args)
{
    String S = "geeksforgeeks";
 
    System.out.println(minCost(S));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function that return true if the
# given character is a vowel
def isVowel(ch):
 
    if (ch == 'a' or ch == 'e' or
        ch == 'i' or ch == 'o' or
        ch == 'u'):
        return True;
    else:
        return False;
 
# Function to return the minimum
# cost to convert all the vowels
# of a string to a single one
def minCost(S):
 
    # Stores count of
    # respective vowels
    cA = 0;
    cE = 0;
    cI = 0;
    cO = 0;
    cU = 0;
 
    # Iterate through the string
    for i in range(len(S)):
 
        # If a vowel is encountered
        if (isVowel(S[i])):
 
            # Calculate the cost
            cA += abs(ord(S[i]) - ord('a'));
            cE += abs(ord(S[i]) - ord('e'));
            cI += abs(ord(S[i]) - ord('i'));
            cO += abs(ord(S[i]) - ord('o'));
            cU += abs(ord(S[i]) - ord('u'));
         
    # Return the minimum cost
    return min(min(min(min(cA, cE), cI), cO), cU);
 
# Driver code
S = "geeksforgeeks";
 
print(minCost(S))
 
# This code is contributed by grand_master

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function that return true if the
// given character is a vowel
static bool isVowel(char ch)
{
    if (ch == 'a' || ch == 'e' ||
        ch == 'i' || ch == 'o' ||
        ch == 'u')
        return true;
    else
        return false;
}
 
// Function to return the minimum
// cost to convert all the vowels
// of a string to a single one
static int minCost(String S)
{
     
    // Stores count of
    // respective vowels
    int cA = 0;
    int cE = 0;
    int cI = 0;
    int cO = 0;
    int cU = 0;
 
    // Iterate through the string
    for(int i = 0; i < S.Length; i++)
    {
         
        // If a vowel is encountered
        if (isVowel(S[i]))
        {
             
            // Calculate the cost
            cA += Math.Abs(S[i] - 'a');
            cE += Math.Abs(S[i] - 'e');
            cI += Math.Abs(S[i] - 'i');
            cO += Math.Abs(S[i] - 'o');
            cU += Math.Abs(S[i] - 'u');
        }
    }
 
    // Return the minimum cost
    return Math.Min(Math.Min(
           Math.Min(Math.Min(cA, cE),
                       cI), cO), cU);
}
 
// Driver code
public static void Main(String[] args)
{
    String S = "geeksforgeeks";
 
    Console.WriteLine(minCost(S));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program f|| the above approach
 
// Function that return true if the
// given character is a vowel
function isVowel(ch)
{
    if (ch == 'a' || ch == 'e'
        || ch == 'i' || ch == 'o'
        || ch == 'u')
        return true;
    else
        return false;
}
 
// Function to return the minimum
// cost to convert all the vowels
// of a string to a single one
function minCost(S)
{
    // St||es count of
    // respective vowels
    var cA = 0;
    var cE = 0;
    var cI = 0;
    var cO = 0;
    var cU = 0;
 
    // Iterate through the string
    for(var i = 0; i < S.length; i++) {
 
        // If a vowel is encountered
        if (isVowel(S[i])) {
 
            // Calculate the cost
            cA += Math.abs(S.charCodeAt(i) - 'a'.charCodeAt(0));
            cE += Math.abs(S.charCodeAt(i) - 'e'.charCodeAt(0));
            cI += Math.abs(S.charCodeAt(i) - 'i'.charCodeAt(0));
            cO += Math.abs(S.charCodeAt(i) - 'o'.charCodeAt(0));
            cU += Math.abs(S.charCodeAt(i) - 'u'.charCodeAt(0));
        }
    }
 
    // Return the Math.minimum cost
    return Math.min(Math.min(Math.min(Math.min(cA, cE),
                    cI),
                cO),
            cU);
}
 
// Driver Code
var S = "geeksforgeeks";
document.write(minCost(S));
 
// This code is contributed by importantly.

输出：

10

时间复杂度： O(N)
辅助空间： O(1)

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